Problem 58 - College Algebra

Question

Solve each logarithmic equation in Exercises \(49-92 .\) Be sure to reject any value of \(x\) that is not in the domain of the original logarithmic expressions. Give the exact answer. Then, where necessary, use a calculator to obtain a decimal approximation, correct to two decimal places, for the solution. $$ \log _{7}(x+2)=-2 $$

Step-by-Step Solution

Answer

The solution to the logarithmic equation \(\log_{7}(x+2) = -2\) is \(x = -1.98\) (to two decimal places)

Step 1: Understanding the Logarithmic Equation

We have a logarithmic equation, which is \(\log_{7}(x+2) = -2\). This can be rewritten as \(7^{-2} = x+2\). The base of the logarithm is '7' and the number '-2' is the power to which '7' must be raised to produce \(x+2\).

Step 2: Calculating the Power of the Base

Next, calculate the value of 7 raised to the power of -2. This equals \(\frac{1}{7^2}\), or \(\frac{1}{49}\). So the equation becomes \(\frac{1}{49} = x+2\).

Step 3: Solving for \(x\)

Finally, we solve for \(x\). Start by subtracting 2 from both sides of the equation. This results in \(x = \frac{1}{49} -2 = -1.98\), to 2 decimal places.

Step 4: Verifying the Solution

It's crucial to verify if this solution falls within the domain of the original logarithmic equation. The original domain is \(x > -2\), and since -1.98 is greater than -2, it's a valid solution for this logarithmic equation.

Key Concepts

Solving Logarithms

Understanding how to solve logarithmic equations is fundamental to grasping higher-level mathematics. A logarithm equation typically takes the form \( \log_{b}(x) = y \), where \( b \) is the base and the result \( y \) is the power to which the base must be raised to obtain \( x \). The solution strategy involves rewriting the logarithmic equation in its exponential form.

To tackle equations like \( \log_{7}(x+2) = -2 \), we convert the logarithm to its equivalent exponential form. This step is crucial as it simplifies the equation into a more familiar form. In this specific equation, the exponential form is \( 7^{-2} = x+2 \), where \( 7 \) is raised to the power of \( -2 \) to yield \( x+2 \). From this point, it's a matter of isolating \( x \) and solving the equation as you would any other algebraic expression.

Logarithmic Expression Domains

The domain of a logarithmic expression specifies the set of values that \( x \) can take on for the logarithm to be defined. Since a logarithm represents the power to which a number must be raised to yield another number, the argument of a logarithm (the number inside the logarithm) must be positive.

For \( \log_{b}(x) \), the domain is \( x > 0 \), because you can only take the logarithm of a positive number. In our specific equation, \( \log_{7}(x+2) \), the domain is \( x+2 > 0 \) or \( x > -2 \), as the value inside the logarithm (\( x+2 \) in this case) must be greater than zero to be valid. Understanding and identifying the domain is critical, as it can affect the solution. Any potential solution falling outside the domain must be rejected, as it does not satisfy the logarithmic expression.

Decimal Approximations

When solving equations, especially in applied contexts, we often need a decimal approximation to understand the practical implications of the solution. After finding the exact form of \( x \) in an equation like \( \log_{7}(x+2) = -2 \), we sometimes use a calculator to approximate the value to a specific number of decimal places.

For instance, in the given equation, after solving we get \( x \) as a fraction. To make this more useful in practical applications, we seek a decimal approximation. The exact solution, \( \frac{1}{49} - 2 \), can be difficult to interpret, so we round it to two decimal places to get \( -1.98 \). This step often comes after solving for \( x \) and ensures the solution is in a form readily usable in real-world scenarios, such as measurements or financial calculations where decimal precision is required.